Modulus-Argument Form (Edexcel A Level Further Maths: Core Pure): Revision Note

Modulus-argument form

The complex number $z=x+\mathrm{i}y$ is said to be in Cartesian form. There are, however, other ways to write a complex number, such as in modulus-argument (polar) form.

How do I write a complex number in modulus-argument (polar) form?

• The Cartesian form of a complex number, $z=x+\mathrm{i}y$, is written in terms of its real part, $x$, and its imaginary part, $y$

• If we let $r=\left|z\right|$ and $\theta =\arg z$, then it is possible to write a complex number in terms of its modulus, $r$, and its argument, $\theta$, called the modulus-argument (polar) form, given by...

  • $z=r\left(\cos \theta +\mathrm{isin}\theta \right)$

• It is usual to give arguments in the range $-\pi <\theta \le \pi$

  • Negative arguments should be shown clearly, e.g. $z=2\left(\cos \left(-\frac{\pi }{3}\right)+\mathrm{isin}\left(-\frac{\pi }{3}\right)\right)$without simplifying $\cos (-\frac{\pi }{3})$  to either $\cos \left(\frac{\pi }{3}\right)$ or $\frac{1}{2}$

  • Occasionally you could be asked to give arguments in the range $0\le \theta <2\pi$

• If a complex number is given in the form $z=r\left(\cos \theta -\mathrm{isin}\theta \right)$, then it is not currently in modulus-argument (polar) form due to the minus sign, but can be converted as follows…

  • By considering transformations of trigonometric functions, we see that $-\sin \theta \equiv \sin (-\theta )$ and $\cos \theta \equiv \cos (-\theta )$

  • Therefore $z=r\left(\cos \theta -\mathrm{isin}\theta \right)$ can be written as $z=r\left(\cos \left(-\theta \right)+\mathrm{isin}\left(-\theta \right)\right)$, now in the correct form and indicating an argument of $-\theta$

• To convert from modulus-argument (polar) form back to Cartesian form, evaluate the real and imaginary parts

  • E.g. $z=2\left(\cos \left(-\frac{\pi }{3}\right)+\mathrm{isin}\left(-\frac{\pi }{3}\right)\right)$ becomes $z=2\left(\frac{1}{2}+\mathrm{i}\left(-\frac{\sqrt{3}}{2}\right)\right)=1-\sqrt{3} \mathrm{i}$

Operations using modulus-argument form

What are the rules for moduli and arguments under multiplication and division?

• When two complex numbers, $z_1$ and $z_2$, are multiplied to give $z_1{z}_2$, their moduli are also multiplied

  • $\left|z_1{z}_2\right|=\left|z_1\right|\left|z_2\right|$

• When two complex numbers, $z_1$ and $z_2$, are divided to give $\frac{z_1}{z_2}$, their moduli are also divided

  • $\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}$

• When two complex numbers, $z_1$ and $z_2$, are multiplied to give $z_1{z}_2$, their arguments are added

  • $\arg \left(z_1{z}_2\right)=\arg z_1+\arg z_2$

• When two complex numbers, $z_1$and $z_2$, are divided to give $\frac{z_1}{z_2}$, their arguments are subtracted

  • $\arg \left(\frac{z_1}{z_2}\right)=\arg z_1-\arg z_2$

How do I multiply complex numbers in modulus-argument (polar) form?

• The main benefit of writing complex numbers in modulus-argument (polar) form is that they multiply and divide very easily (often quicker than when in Cartesian form)

• To multiply two complex numbers, $z_1$ and $z_2$, in modulus-argument (polar) form we use the rules from above to multiply their moduli and add their arguments

  • $\left|z_1{z}_2\right|=\left|z_1\right|\left|z_2\right|$

  • $\arg \left(z_1{z}_2\right)=\arg z_1+\arg z_2$

• So if $z_1=r_1(\cos {\theta }_1+\mathrm{isin}{\theta }_1)$ and $z_2=r_2(\cos {\theta }_2+\mathrm{isin}{\theta }_2)$ then the rules above give…

  • $z_1{z}_2=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+\mathrm{isin}\left({\theta }_1+{\theta }_2\right)\right)$

• Sometimes the new argument, ${\theta }_1+{\theta }_2$, does not lie in the range $-\pi <\theta \le \pi$ (or  $0\le \theta <2\pi$  if this is being used)

  • An out-of-range argument can be adjusted by either adding or subtracting $2\mathrm{\pi }$

  • E.g. If ${\theta }_1=\frac{2\pi }{3}$ and ${\theta }_2=\frac{\pi }{2}$  then  ${\theta }_1+{\theta }_2=\frac{7\mathrm{\pi }}{6}$ 

    • This is currently not in the range , but by subtracting $2\mathrm{\pi }$ from $\frac{7\mathrm{\pi }}{6}$ to give $-\frac{5\mathrm{\pi }}{6}$, a new argument is formed that lies in the correct range and represents the same angle on an Argand diagram

• The rules of multiplying the moduli and adding the arguments can also be applied when…

  • …multiplying three complex numbers together, $z_1{z}_2{z}_3$, or more

  • …finding powers of a complex number (e.g. $z^2$ can be written as $zz$)

• Whilst not examinable, the rules for multiplication can be proved algebraically by multiplying $z_1=r_1(\cos {\theta }_1+\mathrm{isin}{\theta }_1)$ by $z_2=r_2(\cos {\theta }_2+\mathrm{isin}{\theta }_2)$, expanding the brackets and using compound angle formulae

How do I divide complex numbers in modulus-argument (polar) form?

• To divide two complex numbers, $z_1$ and $z_2$ in modulus-argument (polar) form, we use the rules from above to divide their moduli and subtract their arguments

  • $\left|\frac{z_1}{z_2}\right| =\frac{\left|z_1\right|}{\left|z_2\right|}$

  • $\arg \left(\frac{z_1}{z_2}\right)=\arg z_1-\arg z_2$

• So if $z_1=r_1(\cos {\theta }_1+\mathrm{isin}{\theta }_1)$ and $z_2=r_2(\cos {\theta }_2+\mathrm{isin}{\theta }_2)$ then the rules above give…

  • $\frac{z_1}{z_2}=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+\mathrm{isin}\left({\theta }_1-{\theta }_2\right)\right)$

• As with multiplication, sometimes the new argument, ${\theta }_1-{\theta }_2$, can lie out of the range $-\pi <\theta \le \pi$ (or the range $0<\theta \le 2\pi$ if this is being used)

  • You can add or subtract $2\mathrm{\pi }$ to bring out-of-range arguments back in range

• Whilst not examinable, the rules for division can be proved algebraically by dividing $z_1=r_1(\cos {\theta }_1+\mathrm{isin}{\theta }_1)$ by $z_2=r_2(\cos {\theta }_2+\mathrm{isin}{\theta }_2)$, using complex division and compound angle formulae